suppose the people have weights that are normally distributed with a mean of 177 lb and a standard deviation of 26 lb.
Find the probability that if a person is randomly selected, his weight will be greater than 174 pounds?
Assume that weights of people are normally distributed with a mean of 177 lb and a standard deviation of 26 lb.
Mean = 177
standard deviation = 26
We find z-score using given mean and standard deviation
z = ![\frac{x-mean}{standard deviation}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bx-mean%7D%7Bstandard%20deviation%7D%20%20)
= ![\frac{174-177}{26}](https://tex.z-dn.net/?f=%20%5Cfrac%7B174-177%7D%7B26%7D%20%20)
=-0.11538
Probability (z>-0.11538) = 1 - 0.4562 (use normal distribution table)
= 0.5438
P(weight will be greater than 174 lb) = 0.5438
We are given that
<span>d(p) represents the number of dollars jason gets paid
</span><span>t(p) represents the number of Jason's dollars that go to taxes
</span><span>t(p)=500
Then
d(p) = 500 / 0.25
d(p) = 2000
The value of
</span><span>(d-t)(p) = 2000 - 500 = 1500</span>
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